Question : If $\left(3 y+\frac{3}{y}\right)=8$, then find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Option 1: $5\frac{1}{9}$
Option 2: $4\frac{5}{6}$
Option 3: $7\frac{1}{9}$
Option 4: $9\frac{1}{9}$
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $5\frac{1}{9}$
Solution : Given: $\left(3 y+\frac{3}{y}\right)=8$ $⇒(y+\frac{1}{y})=\frac{8}{3}$ Squaring both sides, $⇒(y^2+\frac{1}{y^2}+2)=\frac{64}{9}$ $⇒(y^2+\frac{1}{y^2})=\frac{64}{9} - 2$ $⇒(y^2+\frac{1}{y^2})=\frac{64-18}{9} $ $⇒(y^2+\frac{1}{y^2})=\frac{46}{9}=5\frac{1}{9}$ Hence, the correct answer is $5\frac{1}{9}$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $\left(4y-\frac{4}{y}\right)=11$, find the value of $\left(y^2+\frac{1}{y^2}\right)$.
Question : If $\left(y^2+\frac{1}{y^2}\right)=74$ and $y>1$, then find the value of $\left(y-\frac{1}{y}\right)$.
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Question : Find the value of the given expression. $\left(2 \frac{1}{2}÷ 1 \frac{7}{8}\right) ÷\left(9 \frac{3}{8}÷ 11 \frac{2}{3} \text { of } \frac{1}{8}\right)$
Question : The value of $\left(\frac{2+7 \times 7 \div 9 \text { of } 9+6 \div 6 \times 2}{4 \div 4 \text { of } 5+7 \times 7 \div 7-6+4}\right)$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile